Accedmychevron_right11thchevron_rightmathchevron_rightPermutations And Combinationschevron_rightExercise 7.1

Questions

  1. Question 3

    Find nn, if (n+4)!=3024n!(n+4)! = 3024 \cdot n!.

  2. Question 4

    If 17!+18!=x9!\dfrac{1}{7!} + \dfrac{1}{8!} = \dfrac{x}{9!}, find xx.

  3. Question 5

    Prove that: (2n+1)!n!=[135(2n1)(2n+1)]2n\dfrac{(2n+1)!}{n!} = [1\cdot 3\cdot 5\cdots(2n-1)(2n+1)]2^n.

  4. Question 6

    Express as a single fraction: (n+2)!(r+2)!+(n+1)!(r+1)!\dfrac{(n+2)!}{(r+2)!} + \dfrac{(n+1)!}{(r+1)!}.

  5. Question 7

    There are four distinct colored balls and four boxes of same colors as those of the balls. Determine the number of possible ways the balls, one each in a box, can be placed such that a ball does not go to a box of its own colour.